# Oscillations, Lagrangian.

1. Jan 16, 2014

### binbagsss

The question is that a mass m of weight mg is attached to a fixed point by a light linear spring of stiffness constant k and natural length a. It is capable of oscillating in a vertical planne. Let θ be the angle of the pendulum wrt to the vertical direction, and r the distance between the mass and the fixed point.

Questions:
1a) The KE of the system?

- First of all I want to be sure I have pictured the set up correctly : Does this system consist of '2 oscillations' - the spring verically up and down and simple pendulum motion?

The answer is : m/2($\dot{r}$)$^{2}$+(m/2)(r$\dot{θ}$) $^{2}$

My Question:

So I know that the KE of the spring oscillations is: m/2($\dot{r}$)$^{2}$
and of the free pendulum is:(m/2)(r$\dot{θ}$) $^{2}$

But , when combining to one term, how can you justify directly adding the terms/ physically explain it. For example, in the second term why is r taken to not vary with time, just as it is in the simple pendulum case?

My free pendulum KE term method:

KE= 1/2m$\dot{s}$$^{2}$ where s is the arc length swept out = rθ. so ds/dt=r$\dot{θ}$+$\dot{r}$θ

- So I can see we have taken dr/dt=0, i.e r taken to be fixed.

1b) This system has 2dof. Does it follow that if the system was reduced to only one of the forms of oscillation - just the 'spring oscillation' or simple pendulum - the system would have only 1 DOF?

2) L=m/2($\dot{n}$$^{2}$)+(n+a)$^{2}$$\dot{θ}$$^{2}$-k/2(n$^{2}$)+mg(n+a)cosθ

where n = r-a*

The questions asks to use a small displacement approximation.This includes to neglect terms higher than the quadratic in displacement:

I don't understand some terms neglected and kept. I think my issue is defining the definition of displacement - for the spring motion, is it just n, but for the simple pendulum motion is it nθ? ( rθ but we have converted to generalized coordinates) *

I then reach the conclusion that I should neglect terms : n$^{2}$ or n$^{2}$θ$^{2}$?

I am not sure how to treat the time derivatives. Is neglecting these terms more significant or an equal footing? My thoughts would be on an equal footing, such that, I should also neglect terms: $\dot{n}$$^{2}$, ($\dot{n}$$\dot{θ}$)$^{2}$

This is the wrong method when I look at the solution. Here are the neglected and kept terms that do not agree with my thoughts above:

Neglected:

n$\dot{θ}$)$^{2}$
- how is this term of order two in displacement? Doesn't it have units length rather than length^2?
$\dot{n}$θ^2 - again I would argue the above.

Kept:
$\dot{n}$$^{2}$
n$^{2}$

Many thanks to anyone who can help shed some light on this, greatly appreciated

2. Jan 16, 2014

### voko

You got the correct result, but the derivation is not correct. $${ds \over dt} \ne r \dot \theta + \dot r \theta$$ but $$\left( {ds \over dt} \right)^2 = \dot r ^2 + r^2 \dot \theta^2 .$$ If you are unsure where the latter equation comes from, you need to review the differential line element in polar coordinates, or more generically, in arbitrary coordinates or in parametric form.

That's pretty much what it means.

What is $a^*$? Is that the equilibrium position of the mass? Can you express it in terms of the stiffness and mass?

It seems you need to review the general theory here. Remember, small oscillations always happen about a stable equilibrium. What can be said of potential energy in stable equilibrium? Specifically, its first and second derivatives with regard to generalized coordinates? What minimal number of terms do you need to retain in the Taylor expansion of potential energy about a position of equilibrium?

3. Jan 17, 2014

### binbagsss

Okay I've had a look at the derivative of (ds/dt)^2, so I was working in the wrong coordinate system and the arc length will not be given by rθ?

But, in terms of the KE of this system, isn't ds the line element swept out, so how is the KE term differing then if the oscillations were only that of a simple pendulum? (Assuming I have pictured the motion correctly - spring oscillations and simple pendulum oscillations?)

Yes, sorry, a is the equilibrium position.

Regarding stable equilibrium and the Taylor expansion wrt to generalized coordinates, the first derivative, being the force, for a conservative system, will vanish.
The second, must be >0 as it is a stable equilibrium.
Typically terms of a higher order can be neglected?

4. Jan 17, 2014

### voko

The arc length will be given by the second formula in my previous post.

I am not sure what is being asked here. If you look at the formula I just mentioned, if the system is confined to move only as a spring, then $\dot \theta = 0$ so, $\dot s^2 = \dot r^2$. If the system is confined to move as a pendulum, then $\dot s^2 = r^2 \dot \theta^2$.

Exactly. The only catch here is that you have two degrees of freedom, so you must retain all the second order terms of the expansion.

5. Jan 18, 2014

### binbagsss

Would the taylor expansion be for L or just the potential energy - I'm sure I've seen both methods in books. But overall I would say all of L?

Also, if it is L, would it be a 4 variable taylor expansion: n, $\dot{n}$,θ,$\dot{θ}$?

6. Jan 18, 2014

### voko

Generally speaking, it should be all of $L$. But the kinetic energy term is treated differently. It is always a sum of terms like $f(x) \dot x^2$, so we just take the zero-th term in the expansion of $f(x)$, ending up with $f(x_0) \dot x^2$.

In case you are wondering why they are treated differently, it is because we retain the lowest order non-constant terms. Which happen to be of order 2 for potential energy.